\section{Normalization and Orthogonality}\label{sec:NormalizaitonAndOrthogonality}
Using the definitions in (\ref{eqn:defR1}) and (\ref{eqn:defR2}), two modes referenced by the indices $i=\{\ell_1,m_1,n_1\}$ and $j=\{\ell_2,m_2,n_2\}$ are defined to be \emph{orthonormal} if the following condition is true,
\begin{align}
\int_0^{z_0}\int_0^{2\pi}\vec{e}_{i}(\rho_0,\phi,z)\times\vec{h}_{j}(\rho_0,\phi,z)\cdot\hat{\rho}\;{\;d\phi}{\;dz}=\delta_{i,j}\label{eqn:defoforthognalitycyl}
\end{align}
where $\rho$ is evaluated on the surface $\rho_0$ and where $\delta_{i,j}$ is the \emph{Kronecker delta} function and is defined as
\begin{align}
\delta_{i,j} = \left\{ \begin{array}{ll}
0 & \textrm{if $i\ne j$}\\
1 & \textrm{if $i=j$}
\end{array} \right.
\end{align}
Note that the metric coefficient $\rho$ is not considered in the equation, but still produces an orthonormal basis. 
Expanding the electric and magnetic basis vectors into $TE_z$ and $TM_z$ components yields the summation of four cases,
\begin{align}
\int_0^{z_0}\int_0^{2\pi}(\vec{e}_{Fi}+\vec{e}_{Ai})\times(\vec{h}_{Fj}+\vec{h}_{Aj})\cdot\hat{\rho}\;d\phi{\;dz}\label{eqn:defoforthognalitycylAF}
&=I_{FF}+I_{AA}+I_{FA}+I_{AF}
\end{align}
where
\begin{align}
I_{FF}&=\int_0^{z_0}\int_0^{2\pi}\vec{e}_{Fi}\times\vec{h}_{Fj}\cdot\hat{\rho}\;d\phi{\;dz}\label{eqn:IFFcyl}\\
I_{AA}&=\int_0^{z_0}\int_0^{2\pi}\vec{e}_{Ai}\times\vec{h}_{Aj}\cdot\hat{\rho}\;d\phi{\;dz}\label{eqn:IAAcyl}\\
I_{FA}&=\int_0^{z_0}\int_0^{2\pi}\vec{e}_{Ai}\times\vec{h}_{Fj}\cdot\hat{\rho}\;d\phi{\;dz}\label{eqn:IFAcyl}\\
I_{AF}&=\int_0^{z_0}\int_0^{2\pi}\vec{e}_{Fi}\times\vec{h}_{Aj}\cdot\hat{\rho}\;d\phi{\;dz}\label{eqn:IAFcyl}
\end{align}

Each of these four integrals will be analyzed separately.
\subsection{Case 1:  $TE_z\Leftrightarrow TE_z$}
Expanding the integrand and substituting (\ref{eqn:EFbasiscyl}) and (\ref{eqn:HFbasiscyl}) into (\ref{eqn:IFFcyl}) gives
\begin{multline}\label{eqn:IFFcyl2}
I_{FF}=\int_0^{z_0}\int_0^{2\pi}e_{F\phi i}h_{Fzj}\;d\phi{\;dz}=\\-C_{f\ell_1,m_1,n_1}C_{f\ell_2,m_2,n_2}k_{f\rho{n_1}}k_{f\rho{n_2}}\sqrt{\frac{Z_{f{n_1}}}{Z_{f{n_2}}}}I_{\Phi FF}I_{ZFF}
\end{multline}
where,
\begin{align}
I_{\Phi FF} &= \int_0^{2\pi}\Phi_F(\phi)_{\ell_1,m_1}\Phi_F(\phi)_{\ell_2,m_2}{\;d\phi}\label{eqn:IPhicyl}\\
&=\int_0^{2\pi}\cos\left(m_1\phi-\ell_1\frac{\pi}{2}\right)\cos\left(m_2\phi-\ell_2\frac{\pi}{2}\right){\;d\phi}
\end{align}
\begin{align}
I_{ZFF} &= \int_0^{z_0}Z_F(z)_{n_1}Z_F(z)_{n_2}{\;dz}\label{eqn:IZcylTE}\\
&=\int_0^{z_0}\sin\left(\frac{n_1\pi}{z_0}z\right)\sin\left(\frac{n_2\pi}{z_0}z\right){\;dz}
\end{align}
When $m_1\ne m_2$ and for any combination of $\ell_1$ and $\ell_2$,
\begin{align}
I_{\Phi FF}&=0 \label{eqn:IPHIcylequal0}
\end{align}
When $n_1\ne n_2$ then,
\begin{align}
I_{ZFF}&=0
\end{align}
which proves that the modes are orthogonal. In order to make the modes orthonormal, let $i=j$ so that (\ref{eqn:IFFcyl2}) reduces to,
\begin{align}
-C_{f\ell,m,n}^2k_{f\rho{n}}^2I_{\Phi FF}I_{ZFF}=1\label{eqn:exhcyltemp}
\end{align}
where
\begin{align}
I_{\Phi FF}&=\int_0^{2\pi}\Phi_F^2(\phi)_{\ell,m}{\;d\phi}\\
&=\int_0^{2\pi}\cos^2\left(m\phi-\ell\frac{\pi}{2}\right){\;d\phi}\\
&=\pi\left[1+(-1)^\ell\delta_{m,0}\right]\label{eqn:intPhisqr}
\end{align}
and where
\begin{align}
I_{ZFF}&=\int_0^{z_0}Z_F^2(z)_{n}{\;dz}\\
&=\int_0^{z_0}\sin^2\left(\frac{n\pi}{z_0}z\right){\;dz}\\
&=\left(1-\delta_{n,0}\right)\frac{z_0}{2}\label{eqn:intZsqr}
\end{align}
Finally solving for $C_f$ in (\ref{eqn:exhcyltemp}) and substituting (\ref{eqn:intPhisqr}) and (\ref{eqn:intZsqr}) yields,
\begin{align}
C_{f\ell,m,n} = \left[-{z_0}k_{f\rho{n}}^2\frac{\pi}{2}\left[1+(-1)^\ell\delta_{m,0}\right]\left[1-\delta_{n,0}\right]\right]^{-\frac{1}{2}}
\end{align}

\subsection{Case 2:  $TM_z\Leftrightarrow TM_z$}
\begin{align}
I_{AA}&=\int_0^{z_0}\int_0^{2\pi}\vec{e}_{Ai}\times\vec{h}_{Aj}\cdot\hat{\rho}\;d\phi{\;dz}\\&=
\int_0^{z_0}\int_0^{2\pi}-e_{Azi}h_{A\phi j}\;d\phi{\;dz}\label{eqn:IAA}\\
&=C_{a\ell_1,m_1,n_1}C_{a\ell_2,m_2,n_2}k_{a\rho{n_1}}k_{a\rho{n_2}}\sqrt{\frac{Z_{a{n_1}}}{Z_{a{n_2}}}}I_{\Phi AA}I_{ZAA}
\end{align}
Expanding the integrand and substituting (\ref{eqn:EAbasiscyl}) and (\ref{eqn:HAbasiscyl}) into (\ref{eqn:IAAcyl}) gives
\begin{multline}
\int_0^{z_0}\int_0^{2\pi}-e_{Azi}h_{A\phi j}\;({\rho_0}{\;d\phi}){\;dz}\\=-C_{a\ell_1,m_1,n_1}C_{a\ell_2,m_2,n_2}k_{a\rho{n_1}}k_{a\rho{n_2}}\sqrt{\frac{Z_{a{n_1}}}{Z_{a{n_2}}}}I_{\Phi AA}I_{ZAA}
\end{multline}
where $I_{\Phi AA}$ is given in (\ref{eqn:IPhicyl})
\begin{align}
I_{\Phi AA} &= \int_0^{2\pi}\Phi_A(\phi)_{\ell_1,m_1}\Phi_A(\phi)_{\ell_2,m_2}{\;d\phi}\label{eqn:IPhicylA}\\
&=\int_0^{2\pi}\cos\left(m_1\phi-\ell_1\frac{\pi}{2}\right)\cos\left(m_2\phi-\ell_2\frac{\pi}{2}\right){\;d\phi}
\end{align}
and,
\begin{align}
I_{ZAA} &= \int_0^{z_0}Z_A(z)_{n_1}Z_A(z)_{n_2}{\;dz}\label{eqn:IZcylTM}\\
&=\int_0^{z_0}\cos\left(\frac{n_1\pi}{z_0}z\right)\cos\left(\frac{n_2\pi}{z_0}z\right){\;dz}
\end{align}
When $m_1\ne m_2$ and for any combination of $\ell_1$ and $\ell_2$ then $I_{\Phi AA}=0$. When $n_1\ne n_2$ then $I_{ZAA}=0$
which proves that the modes are orthogonal. In order to make the modes orthonormal, let $i=j$ so that (\ref{eqn:IAAcyl}) reduces to,
\begin{align}
-C_{a\ell,m,n}^2k_{a\rho{n}}^2I_{\Phi AA}I_{ZAA}
=1\label{eqn:exhcyltemp2}
\end{align}
where $I_{\Phi AA}$ reduces to,
\begin{align}
I_{\Phi AA} &= \int_0^{2\pi}\Phi^2_A(\phi)_{\ell,m}{\;d\phi}\\
&=\int_0^{2\pi}\cos^2\left(m\phi-\ell\frac{\pi}{2}\right){\;d\phi}\\
&=\pi\left[1+(-1)^\ell\delta_{m,0}\right]\label{eqn:intPhisqrA}
\end{align}
and 
\begin{align}
I_{ZAA}&=\int_0^{z_0}Z^2_A(z)_{n}{\;dz}\\
&=\int_0^{z_0}\cos^2\left(\frac{n\pi}{z_0}z\right){\;dz}\\
&=\left(1+\delta_{n,0}\right)\frac{z_0}{2}\label{eqn:intZsqrTM}
\end{align}
Finally solving for $C_a$ in (\ref{eqn:exhcyltemp2}) and substituting (\ref{eqn:intPhisqrA}) and (\ref{eqn:intZsqrTM}) yields,
\begin{align}
C_{a\ell,m,n} = \left[-{z_0}k_{a\rho{n}}^2\frac{\pi}{2}\left[1+(-1)^\ell\delta_{m,0}\right]\left[1+\delta_{n,0}\right]\right]^{-\frac{1}{2}}
\end{align}

\subsection{Case 3:  $TE_z\Leftrightarrow TM_z$}
Expanding the cross product of the integrand in (\ref{eqn:IFAcyl}) yields,
\begin{align}
I_{FA}&=\int_0^{z_0}\int_0^{2\pi}\vec{e}_{Ai}\times\vec{h}_{Fj}\cdot\hat{\rho}\;d\phi{\;dz}=0\label{eqn:IAF}
\end{align}

\subsection{Case 4:  $TM_z\Leftrightarrow TE_z$}
\begin{align}
I_{AF}&=\int_0^{z_0}\int_0^{2\pi}\vec{e}_{Fi}\times\vec{h}_{Aj}\cdot\hat{\rho}\;d\phi{\;dz}\\
&=\int_0^{z_0}\int_0^{2\pi}(e_{A\phi i}h_{Fzj}-e_{Azi}h_{F\phi j})\;d\phi{\;dz}\label{eqn:IFA}
\end{align}
Expanding the integrand and substituting (\ref{eqn:EFbasiscyl}), (\ref{eqn:HFbasiscyl}), (\ref{eqn:EAbasiscyl}) and (\ref{eqn:HAbasiscyl}) yields,
\begin{align}
I_{AF}&=C_{a\ell_2,m_2,n_2}C_{f\ell_1,m_1,n_1}\frac{\sqrt{Z_{an_2}}}{\sqrt{Z_{fn_1}}}\left[\frac{k_{f\rho n_1}}{k_{a\rho n_2}}I_{\Phi FA}I_{ZFA}-\frac{k_{a\rho n_2}}{k_{f\rho n_1}}I_{\Phi AF}I_{ZAF}\right]\label{eqn:IAFcyl2}
\end{align}
where
\begin{align}
I_{\Phi FA}&=\int_0^{2\pi}\Phi_F(\phi)\Phi'_A(\phi)\;d\phi\\
&=-m_2\int_0^{2\pi}\cos\left(m_1\phi-\ell_1\frac{\pi}{2}\right)\sin\left(m_2\phi-\ell_2\frac{\pi}{2}\right)\;d\phi\\
I_{ZFA}&=\int_0^{z_0}Z_F(z)Z'_A(z)\;dz\\
&=-\frac{n_2\pi}{z_0}\int_0^{z_0}\sin\left(\frac{n_1\pi}{z_0}z\right)\sin\left(\frac{n_2\pi}{z_0}z\right)\;dz\\
I_{\Phi AF}&=\int_0^{2\pi}\Phi_A(\phi)\Phi'_F(\phi)\;d\phi\\
&=-m_1\int_0^{2\pi}\cos\left(m_2\phi-\ell_2\frac{\pi}{2}\right)\sin\left(m_1\phi-\ell_1\frac{\pi}{2}\right)\;d\phi\\
I_{ZAF}&=\int_0^{z_0}Z_A(z)Z'_F(z)\;dz\\
&=\frac{n_1\pi}{z_0}\int_0^{z_0}\cos\left(\frac{n_2\pi}{z_0}z\right)\cos\left(\frac{n_1\pi}{z_0}z\right)\;dz
\end{align}
From the above integrals it is found that if $m_1 \ne m_2$ and $n_1 \ne n_2$ then $I_{AF}=0$. Also if
$m_1=0$, $m_2=0$, $n_1=0$ or $n_2=0$ then $I_{AF}=0$. Under the conditions of $m_1=m_2$ and $n_1=n_2$ then from (\ref{eqn:IAFcyl2}) it is seen that 
$$
\frac{k_{f\rho n}}{k_{a\rho n}}=\frac{k_{a\rho n}}{k_{f\rho n}}=1
$$
Also under the above conditions and $\ell_1 \ne \ell_2$ or $\ell_1=1-\ell_2$ then,
\begin{align}
I_{\Phi{FA}}&=-m\int_0^{2\pi}\cos\left(m\phi-\ell_1\frac{\pi}{2}\right)\sin\left(m\phi-\ell_2\frac{\pi}{2}\right)\;d\phi=m\pi(-1)^{\ell_1}\\
I_{ZFA}&=-\frac{n\pi}{z_0}\int_0^{z_0}\sin^2\left(\frac{n\pi}{z_0}z\right)\;dz=-\frac{n\pi}{2}\left(1-\delta_{n,0}\right)\\
I_{\Phi{AF}}&=-m\int_0^{2\pi}\cos\left(m\phi-\ell_2\frac{\pi}{2}\right)\sin\left(m\phi-\ell_1\frac{\pi}{2}\right)\;d\phi=m\pi(-1)^{\ell_2}\\
I_{ZAF}&=\frac{n\pi}{z_0}\int_0^{z_0}\cos^2\left(\frac{n\pi}{z_0}z\right)\;dz=\frac{n\pi}{2}\left(1+\delta_{n,0}\right)
\end{align}
If $\ell_1=\ell_2$ then $I_{\Phi{FA}}=I_{\Phi{AF}}=0$. Finally, substituting the above results into (\ref{eqn:IAFcyl2}) yields,
\begin{align}
I_{AF}&=-C_{a\ell_2,m_2,n_2}C_{f\ell_1,m_1,n}\frac{\sqrt{Z_{an}}}{\sqrt{Z_{fn_1}}}\frac{mn\pi^2}{2}\left[(-1)^{1-\ell}+(-1)^{\ell}\right]=0\label{eqn:IAFcyl3}
\end{align}
where $\ell=0,1$.  The above results from case 3 and case 4 show that $TE_z$ and $TM_z$ basis modes are orthogonal and therefore can be considered as distinct modes in the basis expansion given that they satisfy all boundary conditions.


\section{Power}
The total power is given by the \emph{Poynting theorem} as,
\begin{align}
P &= \frac{1}{2}\Re\left\{\iint_S\vec{E}\times\vec{H}^*\cdot d\vec{s}\right\}\\
&= \frac{1}{2}\Re\left\{\int_0^{2\pi}\int_0^{z_0}\vec{E}_\bot\times\vec{H}_\bot^*\cdot\hat{\rho}\; (\rho_0 d\phi)dz\right\}
\end{align}
where $\rho = \rho_0$ and where $^*$ is the complex conjugate operator. Expanding the electric and magnetic fields into $TE_z$ and $TM_z$ components yields the summation of four cases,
\begin{align}
\int_0^{z_0}\int_0^{2\pi}(\vec{E}_{F}+\vec{E}_{A})\times(\vec{H}^*_{F}+\vec{H}^*_{A})\cdot\hat{\rho_0}\;({\rho}{\;d\phi}){\;dz}\label{eqn:defoforthognalitycylAFpwr}
&=P_{F}+P_{A}+P_{FA}+P_{AF}
\end{align}
where
\begin{align}
P_{F}&=\frac{1}{2}\Re\left\{\int_0^{z_0}\int_0^{2\pi}\vec{E}_{F}\times\vec{H}^*_{F}\cdot\hat{\rho}\;({\rho_0}{\;d\phi}){\;dz}\right\}\label{eqn:IFFcylpwr}\\
P_{A}&=\frac{1}{2}\Re\left\{\int_0^{z_0}\int_0^{2\pi}\vec{E}_{A}\times\vec{H}^*_{A}\cdot\hat{\rho}\;({\rho_0}{\;d\phi}){\;dz}\right\}\label{eqn:IAAcylpwr}\\
P_{FA}&=\frac{1}{2}\Re\left\{\int_0^{z_0}\int_0^{2\pi}\vec{E}_{F}\times\vec{H}^*_{A}\cdot\hat{\rho}\;({\rho_0}{\;d\phi}){\;dz}\right\}\label{eqn:IFAcylpwr}\\
P_{AF}&=\frac{1}{2}\Re\left\{\int_0^{z_0}\int_0^{2\pi}\vec{E}_{A}\times\vec{H}^*_{F}\cdot\hat{\rho}\;({\rho_0}{\;d\phi}){\;dz}\right\}\label{eqn:IAFcylpwr}
\end{align}
From section \ref{sec:NormalizaitonAndOrthogonality} it was shown that $TE_z$ and $TM_z$ fields are orthogonal and therefore $P_{FA}=P_{AF}=0$.  This means that the total power is a summation of the power in the $TE_z$ fields and the power in the $TM_z$ fields.  
\begin{align}
P = P_{F} + P_{A}
\end{align}
It was also shown that each field was expanded as a summation of $TE_z$ modes and $TM_z$ modes where individual modes were orthogonal to each other. This means that the total power is equal to the summation of the power in each individual mode.
\begin{align}
P = \sum_{i}p_{Fi} + \sum_{j}p_{Aj}
\end{align}
where $P_{F}=\sum p_{Fi}$ and $P_{A}=\sum p_{Aj}$.
The individual index variables $i=\{\ell_f,m_f,n_f\}$ and $j=\{\ell_a,m_a,n_a\}$ represent a triple of index variables and when used in conjunction with a summation a triple summation is assumed.

\subsection{Power $TE_z$}

Substituting (\ref{eqn:EFperpcyl}) and (\ref{eqn:HFperpcyl}) yields,
\begin{align}
P_{F} &= \frac{1}{2}\Re\left\{\sum_{i}\frac{R'_F(\rho_0)_{i}}{-k_{f\rho i}}R_F^*(\rho_0)_{i}\int_0^{z_0}\int_0^{2\pi}\vec{e}_F(\phi,z)_{i}\times\vec{h}^*_F(\phi,z)_{i}\cdot{\hat{\rho}}\;(\rho_0 d\phi)dz\right\}\\
&=\frac{1}{2}\Re\left\{\sum_{i}\frac{R'_F(\rho_0)_{i}}{-k_{f\rho i}}R_F^*(\rho_0)_{i}\int_0^{z_0}\int_0^{2\pi}{e}_{F\phi}(\phi,z)_{i}{h}^*_{Fz}(\phi,z)_{i}\;(\rho_0 d\phi)dz\right\}
\end{align}
Substituting the basis vectors (\ref{eqn:EFbasiscyl}) and (\ref{eqn:HFbasiscyl}) yields,
\begin{align}
P_F &= \frac{1}{2}\Re\left\{\sum_{i}-\rho_0 C_{fi}C^*_{fi}k_{f\rho{i}}k^*_{f\rho{i}}\frac{Z_{fi}}{|Z_{fi}|}\int_0^{2\pi}|\Phi(\phi)_{\ell,m}|^2\;d\phi \int_0^{z_0}|Z(z)_n|^2\;dz\frac{R'_F(\rho_0)_{i}}{-k_{f\rho i}}R^*_F(\rho_0)_{i}\right\}\label{eqn:Pwrcyl}
\end{align}
Since $\Phi(\phi)_{\ell,m}$ and $Z(z)_n$ are real valued functions then $|\Phi(\phi)_{\ell,m}|^2=\Phi(\phi)^2_{\ell,m}$ and
$|Z(z)_n|^2=Z(z)^2_n$. Substituting into \ref{eqn:Pwrcyl} yields, 
\begin{align}
P_F &= \frac{1}{2}\Re\left\{\sum_{i}-\rho_0 C_{fi}C^*_{fi}k_{f\rho{i}}k^*_{f\rho{i}}\frac{Z_{fi}}{|Z_{fi}|} I_{\Phi FF} I_{ZFF} \frac{R'_F(\rho_0)_{i}}{-k_{f\rho i}}R^*_F(\rho_0)_{i}\right\}\label{eqn:Pwrcyl1_5}
\end{align}
where $I_{\Phi FF}$ and $I_{ZFF}$ are defined in (\ref{eqn:intPhisqr}) and (\ref{eqn:intZsqr}). From (\ref{eqn:exhcyltemp}) it is found that,
\begin{align}
I_{\Phi FF}I_{ZFF}=\frac{1}{-C_{fi}^2k_{f\rho{i}}^2}\label{eqn:Pwrcyl1_5_5}
\end{align}
and then substituting,
\begin{align}
P_F=\frac{1}{2}\Re\left\{\sum_{i}\frac{C^*_{fi}}{C_{fi}}\frac{k^*_{f\rho{i}}}{k_{f\rho{i}}}\frac{Z_{fi}}{|Z_{fi}|}\rho_0 \frac{R'_F(\rho_0)_{i}}{-k_{f\rho{i}}} R_F^*(\rho_0)_{i}\right\}\label{eqn:Pwrcyl2}
\end{align}
Since $I_{\Phi FF}>0$, $I_{ZFF}>0$ and $\rho>0$ it can be shown from (\ref{eqn:Pwrcyl1_5_5}) that, 
\begin{align}
\frac{C^*_{fi}}{C_{fi}}\frac{k^*_{f\rho{i}}}{k_{f\rho{i}}}=-1
\end{align}
then the total power power is given by
\begin{align}
P_F&=\frac{1}{2}\Re\left\{\sum_{i}\rho_0\frac{R'_F(\rho_0)_{i}}{-k_{f\rho{i}}}R_F^*(\rho_0)_{i}\frac{Z_{fi}}{|Z_{fi}|}\right\}\label{eqn:Pwrcyl3}
\end{align}
It can be seen that the the total power is a summation of the power of each individual mode.

\subsection{Power $TM_z$}
%The total power is given by the \emph{Poynting theorem} in (\ref{eqn:PoyntingThereom}) and (\ref{eqn:PoyntingVector}).
Substituting (\ref{eqn:EAperpcyl}) and (\ref{eqn:EAperpcyl}) yields,
\begin{align}
\vec{P}_A &= \frac{1}{2}\Re\left\{\sum_jR_A(\rho_0)_{j}\frac{R'^*_A(\rho_0)_j}{-k^*_{a\rho j}}\int_0^{z_0}\int_0^{2\pi}\vec{e}_A(\phi,z)_{i}\times\vec{h}^*_A(\phi,z)_{i}\cdot{\hat{\rho}}\;(\rho_0 d\phi)dz\right\}\label{eqn:PoyntingVectorTM2}\\
&= \frac{1}{2}\Re\left\{\sum_jR_A(\rho_0)_{j}\frac{R'^*_A(\rho_0)_j}{-k^*_{a\rho j}}\int_0^{z_0}\int_0^{2\pi}-{e}_{Az}(\phi,z)_{i}{h}^*_{A \phi}(\phi,z)_{i}\;(\rho_0 d\phi)dz\right\}\label{eqn:PoyntingVectorTM3}
\end{align}
Substituting the basis vectors (\ref{eqn:EAbasiscyl}) and (\ref{eqn:HAbasiscyl}) yields,
\begin{align}
P_A &= \frac{1}{2}\Re\left\{\sum_{j}R_A(\rho_0)_{j}\frac{R'^*_A(\rho_0)_{j}}{-k^*_{a\rho j}}C_{aj}C^*_{aj}k_{a\rho{j}}k^*_{a\rho{j}}\frac{Z_{ai}}{|Z_{ai}|}\rho_0 I_{\Phi AA}I_{ZAA}\right\}\label{eqn:PwrcylTM}
\end{align}
where $I_{\Phi AA}$ and $I_{ZAA}$ are defined in (\ref{eqn:intPhisqrA}) and (\ref{eqn:intZsqrTM}). From (\ref{eqn:exhcyltemp2}) it is found that,
\begin{align}
I_{\Phi AA}I_{ZAA}=\frac{1}{C_{aj}^2k_{a\rho j}^2}
\end{align}
and then substituting,
\begin{align}
P_A=\frac{1}{2}\Re\left\{\sum_{j}\rho_0 R_A(\rho_0)_j\frac{R'^*_A(\rho_0)_j}{-k^*_{a\rho{j}}}\frac{C^*_{aj}}{C_{aj}}\frac{k^*_{a\rho{j}}}{k_{a\rho{j}}}\frac{Z_{ai}}{|Z_{ai}|}\right\}\label{eqn:Pwrcyl2TM}
\end{align}
Since,
\begin{align}
\frac{C^*_{aj}}{C_{aj}}\frac{k^*_{a\rho{j}}}{k_{a\rho{j}}}=1
\end{align}
then the total power power of the $TM_z$ modes is given by
\begin{align}
P_A&=\frac{1}{2}\Re\left\{\sum_{j}\rho_0 R_A(\rho_0)_j\frac{R'^*_A(\rho_0)_j}{-k^*_{a\rho{j}}}\frac{Z_{ai}}{|Z_{ai}|}\right\}\label{eqn:Pwrcyl3TM}\\
\end{align}

The total power is found by summing (\ref{eqn:Pwrcyl3}) and (\ref{eqn:Pwrcyl3TM}) to get
\begin{align}
P&=\frac{1}{2}\Re\left\{\sum_{i}\rho_0 \frac{R'_F(\rho_0)_{i}}{-k_{f\rho{i}}}R_F^*(\rho_0)_{i}\frac{Z_{fi}}{|Z_{fi}|}+\sum_{j}\rho_0 R_A(\rho_0)_j\frac{R'^*_A(\rho_0)_j}{-k^*_{a\rho{j}}}\frac{Z_{ai}}{|Z_{ai}|}\right\}\label{eqn:Pwrtotalcyl}
\end{align}
